I\'m trying to use the python-docx module to replace a word in a file and save the new file with the caveat that the new file must have exactly the same formatting as the ol
I've forked a repo of python-docx here, which preserves all of the preexisting data in a docx file, including formatting. Hopefully this is what you're looking for.
In addition to @ramil, you have to escape some characters before placing them as string values into the XML, so this worked for me:
def escape(escapee):
escapee = escapee.replace("&", "&")
escapee = escapee.replace("<", "<")
escapee = escapee.replace(">", ">")
escapee = escapee.replace("\"", """)
escapee = escapee.replace("'", "'")
return escapee
from docx import Document
file_path = 'C:/tmp.docx'
document = Document(file_path)
def docx_replace(doc_obj, data: dict):
"""example: data=dict(order_id=123), result: {order_id} -> 123"""
for paragraph in doc_obj.paragraphs:
for key, val in data.items():
key_name = '{{{}}}'.format(key)
if key_name in paragraph.text:
paragraph.text = paragraph.text.replace(key_name, str(val))
for table in doc_obj.tables:
for row in table.rows:
for cell in row.cells:
docx_replace(cell, data)
docx_replace(document, dict(order_id=123, year=2018, payer_fio='payer_fio', payer_fio1='payer_fio1'))
document.save(file_path)
The problem with the methods above is that they lose the existing formatting. Please see my answer which performs the replace and retains formatting.
There is also python-docx-template
which allows jinja2 style templating within a docx template. Here's a link to the documentation
this worked for me:
def docx_replace(old_file,new_file,rep):
zin = zipfile.ZipFile (old_file, 'r')
zout = zipfile.ZipFile (new_file, 'w')
for item in zin.infolist():
buffer = zin.read(item.filename)
if (item.filename == 'word/document.xml'):
res = buffer.decode("utf-8")
for r in rep:
res = res.replace(r,rep[r])
buffer = res.encode("utf-8")
zout.writestr(item, buffer)
zout.close()
zin.close()
As it seems to be, Docx for Python is not meant to store a full Docx with images, headers, ... , but only contains the inner content of the document. So there's no simple way to do this.
Howewer, here is how you could do it:
First, have a look at the docx tag wiki:
It explains how the docx file can be unzipped: Here's how a typical file looks like:
+--docProps
| + app.xml
| \ core.xml
+ res.log
+--word //this folder contains most of the files that control the content of the document
| + document.xml //Is the actual content of the document
| + endnotes.xml
| + fontTable.xml
| + footer1.xml //Containst the elements in the footer of the document
| + footnotes.xml
| +--media //This folder contains all images embedded in the word
| | \ image1.jpeg
| + settings.xml
| + styles.xml
| + stylesWithEffects.xml
| +--theme
| | \ theme1.xml
| + webSettings.xml
| \--_rels
| \ document.xml.rels //this document tells word where the images are situated
+ [Content_Types].xml
\--_rels
\ .rels
Docx only gets one part of the document, in the method opendocx
def opendocx(file):
'''Open a docx file, return a document XML tree'''
mydoc = zipfile.ZipFile(file)
xmlcontent = mydoc.read('word/document.xml')
document = etree.fromstring(xmlcontent)
return document
It only gets the document.xml file.
What I recommend you to do is:
If you have node.js installed, be informed that I have worked on DocxGenJS which is templating engine for docx documents, the library is in active development and will be released soon as a node module.